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Prove that if $m$ satisfy inequality: $$\left (1+ \frac{1}{2m}\right) \log_k 3 - \log_k (27 + 3^\frac{1}{2m}) \le 2$$ where $k=\frac{1}{2}$, then $x^2 + mx + 1 > 0$ for every real number. I solved one $m_1=0$, but the other doesn't satisfy the second condition.

Bernard M.
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1 Answers1

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HINT.-You have $$\left (1+ \frac{1}{2m}\right) \log_k x - \log_k (27 + 3^\frac{1}{2m}) \le 2\iff\log_k\frac{x^{1+\frac{1}{2m}}}{27+ 3^{\frac{1}{2m}}}\le2=\log_k\frac14$$ and because the log is injective and increasing we get $$\frac{x^{1+\frac{1}{2m}}}{27+ 3^{\frac{1}{2m}}}\lt\frac14$$ Now you can work in an easier way.Try it and if you can't finish I shall give you the final part.

Piquito
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  • Gosh I'm sorry, I made a mistake. $x$ should be just $3$. Btw I got to the point you're talking about, but solving in farther is getting me to one $m_2$ which doesn't satisfy the second condition – Bernard M. Oct 20 '16 at 15:39
  • With $3$ instead of $x$ the problem becomes easier to answer. I don't need anymore the final part I expected give you. Regards. – Piquito Oct 20 '16 at 15:53
  • It's my homework for tommorrow, I can't see how it's easier now. Could help me more? :/ – Bernard M. Oct 20 '16 at 16:07